Can Python test the membership of multiple values in a list

Question

I want to test if two or more values have membership on a list  but I m getting an unexpected result    gt  gt  gt   a   b  in   b    a    foo    bar     a   True    So  Can Python test the membership of multiple values at once in a list  What does that result mean

User · Answer

I m pretty sure in is having higher precedence than   so your statement is being interpreted as  a     b  in   b         which then evaluates to  a   True since  b  is in the array   See previous answer for how to do what you want

User · Answer

The Python parser evaluated that statement as a tuple  where the first value was  a   and the second value is the expression  b  in   b    a    foo    bar    which evaluates to True    You can write a simple function do do what you want  though   def all in candidates  sequence       for element in candidates          if element not in sequence              return False     return True   And call it like    gt  gt  gt  all in   a    b      b    a    foo    bar    True

User · Answer

This does what you want  and will work in nearly all cases    gt  gt  gt  all x in   b    a    foo    bar   for x in   a    b    True   The expression  a   b  in   b    a    foo    bar   doesn t work as expected because Python interprets it as a tuple    gt  gt  gt   a    b    a    b    gt  gt  gt   a   5   2   a   7   gt  gt  gt   a    x  in  xerxes    a   True    Other Options  There are other ways to execute this test  but they won t work for as many different kinds of inputs  As Kabie points out  you can solve this problem using sets      gt  gt  gt  set   a    b    issubset set   a    b    foo    bar     True  gt  gt  gt    a    b    lt     a    b    foo    bar   True      sometimes    gt  gt  gt    a     b     lt     a     b     foo    bar   Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt  TypeError  unhashable type   list    Sets can only be created with hashable elements  But the generator expression all x in container for x in items  can handle almost any container type  The only requirement is that container be re-iterable  i e  not a generator   items can be any iterable at all    gt  gt  gt  container      b     a    foo    bar    gt  gt  gt  items    i for i in   a     b      gt  gt  gt  all x in    b     a    foo    bar   for x in items  True   Speed Tests  In many cases  the subset test will be faster than all  but the difference isn t shocking -- except when the question is irrelevant because sets aren t an option  Converting lists to sets just for the purpose of a test like this won t always be worth the trouble  And converting generators to sets can sometimes be incredibly wasteful  slowing programs down by many orders of magnitude   Here are a few benchmarks for illustration  The biggest difference comes when both container and items are relatively small  In that case  the subset approach is about an order of magnitude faster    gt  gt  gt  smallset   set range 10    gt  gt  gt  smallsubset   set range 5    gt  gt  gt   timeit smallset  gt   smallsubset 110 ns    0 702 ns per loop  mean    std  dev  of 7 runs  10000000 loops each   gt  gt  gt   timeit all x in smallset for x in smallsubset  951 ns    11 5 ns per loop  mean    std  dev  of 7 runs  1000000 loops each    This looks like a big difference  But as long as container is a set  all is still perfectly usable at vastly larger scales    gt  gt  gt  bigset   set range 100000    gt  gt  gt  bigsubset   set range 50000    gt  gt  gt   timeit bigset  gt   bigsubset 1 14 ms    13 9   s per loop  mean    std  dev  of 7 runs  1000 loops each   gt  gt  gt   timeit all x in bigset for x in bigsubset  5 96 ms    37   s per loop  mean    std  dev  of 7 runs  100 loops each    Using subset testing is still faster  but only by about 5x at this scale  The speed boost is due to Python s fast c-backed implementation of set  but the fundamental algorithm is the same in both cases   If your items are already stored in a list for other reasons  then you ll have to convert them to a set before using the subset test approach  Then the speedup drops to about 2 5x    gt  gt  gt   timeit bigset  gt   set bigsubseq  2 1 ms    49 2   s per loop  mean    std  dev  of 7 runs  100 loops each    And if your container is a sequence  and needs to be converted first  then the speedup is even smaller    gt  gt  gt   timeit set bigseq   gt   set bigsubseq  4 36 ms    31 4   s per loop  mean    std  dev  of 7 runs  100 loops each    The only time we get disastrously slow results is when we leave container as a sequence    gt  gt  gt   timeit all x in bigseq for x in bigsubseq  184 ms    994   s per loop  mean    std  dev  of 7 runs  10 loops each    And of course  we ll only do that if we must  If all the items in bigseq are hashable  then we ll do this instead    gt  gt  gt   timeit bigset   set bigseq   all x in bigset for x in bigsubseq  7 24 ms    78   s per loop  mean    std  dev  of 7 runs  100 loops each    That s just 1 66x faster than the alternative  set bigseq   gt   set bigsubseq   timed above at 4 36    So subset testing is generally faster  but not by an incredible margin  On the other hand  let s look at when all is faster  What if items is ten-million values long  and is likely to have values that aren t in container    gt  gt  gt   timeit hugeiter    x   10 for bss in  bigsubseq    2000 for x in bss   set bigset   gt   set hugeiter  13 1 s    167 ms per loop  mean    std  dev  of 7 runs  1 loop each   gt  gt  gt   timeit hugeiter    x   10 for bss in  bigsubseq    2000 for x in bss   all x in bigset for x in hugeiter  2 33 ms    65 2   s per loop  mean    std  dev  of 7 runs  100 loops each    Converting the generator into a set turns out to be incredibly wasteful in this case  The set constructor has to consume the entire generator  But the short-circuiting behavior of all ensures that only a small portion of the generator needs to be consumed  so it s faster than a subset test by four orders of magnitude    This is an extreme example  admittedly  But as it shows  you can t assume that one approach or the other will be faster in all cases   The Upshot  Most of the time  converting container to a set is worth it  at least if all its elements are hashable  That s because in for sets is O 1   while in for sequences is O n     On the other hand  using subset testing is probably only worth it sometimes  Definitely do it if your test items are already stored in a set  Otherwise  all is only a little slower  and doesn t require any additional storage  It can also be used with large generators of items  and sometimes provides a massive speedup in that case

User · Answer

how can you be pythonic without lambdas     not to be taken seriously    but this way works too   orig array             test array            filter lambda x x in test array  orig array     test array   leave out the end part if you want to test if any of the values are in the array   filter lambda x x in test array  orig array

User · Answer

x for x in   a   b   if x in   b    a    foo    bar      The reason I think this is better than the chosen answer is that you really don t need to call the  all    function  Empty list evaluates to False in IF statements  non-empty list evaluates to True    if  x for x in   a   b   if x in   b    a    foo    bar            Do something      Example    gt  gt  gt   x for x in   a   b   if x in   b    a    foo    bar      a    b    gt  gt  gt   x for x in   G   F   if x in   b    a    foo    bar

User · Answer

Another way to do it    gt  gt  gt  set   a   b    issubset    b   a   foo   bar     True

User · Answer

Here s how I did it   A     a   b   c   B     c   logic     x in B  for x in A  if True in logic      do something

User · Answer

If you want to check all of your input matches    gt  gt  gt  all x in   b    a    foo    bar   for x in   a    b      if you want to check at least one match    gt  gt  gt  any x in   b    a    foo    bar   for x in   a    b

User · Answer

Both of the answers presented here will not handle repeated elements  For example  if you are testing whether  1 2 2  is a sublist of  1 2 3 4   both will return True  That may be what you mean to do  but I just wanted to clarify  If you want to return false for  1 2 2  in  1 2 3 4   you would need to sort both lists and check each item with a moving index on each list  Just a slightly more complicated for loop

User · Answer

I would say we can even leave those square brackets out    array     b    a    foo    bar   all  i in array for i in  a    b

[python] Can Python test the membership of multiple values in a list?

Examples related to python